# How do you find the asymptotes for #f(x)= (x^2+4x+3)/(x^2 - 9)#?

##### 2 Answers

vertical asymptote=

horizontal asymptote =

#### Explanation:

The **vertical asymptote** is equal to denominator set to zero.Given that your denominator has a x squared the result will be x=

The **horizontal asymptote** in this case is equal to the division of the x with largest exponent in both numerator and denominator .

vertical asymptote x = 3

horizontal asymptote y = 1

#### Explanation:

First step is to factor the function:

#(x^2+4x+3)/(x^2-9) =( (x+3)(x+1))/((x+3)(x-3))# and simplifying :

#(cancel(x+3)(x+1))/(cancel(x+3)(x-3))# left with

# (x+1)/(x-3)# Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation let denominator equal zero.

solve: x - 3 = 0 → x = 3 is equation of asymptote.

Horizontal asymptotes occur as

#lim_(x→±∞) f(x) → 0# now

#(x+1)/(x-3) =( x/x + 1/x)/(x/x -3/x) = (1+1/x)/(1 -3/x) # as x approaches ∞ ,

# 1/x " and " -3/x → 0 # hence y = 1 is the asymptote

here is the graph of the function.

graph{(x^2+4x+3)/(x^2-9) [-10, 10, -5, 5]}